harey

There is a hotel with an infinite number of rooms, all rooms occupied by little green men (one man in a room). An infinity of little blue man arrive and each one asks for a room. No problem, the manager moves the blue man from the room n to the room 2*n, freeing the odd-numbered rooms for the green men. So far, loosely copied from https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel. It turns out that the blue men sing between noon and midnight and sleep between midnight and noon while the green men sleep between noon and midnight and sing between midnight and noon. Complaints. The manager decides to group them. Conveniently, the rooms are in a straight line, numbered from left to right. While there is a green man left to a blue man, he makes them change the rooms. Eventually, all the blue men leave. - how many rooms are free? - how many rooms remain occupied? - what is the number of the first occupied room?

If a coin can be discarded, it does not mean it will be discarded. I would reformulate it: The key question is this: will all coins that are kept at a certain event ever be discarded at a later event? BTW, we can establish a bijection between Al's and Bert's coins. The coins bear green numbers. After each step, Al renumbers them and assigns them blue numbers 1, 3, 5, ... His blue numbers will match the (green) numbers in Bert's box. For any number of steps. I think it is legal to assume it is true even for N-> inf. Another way to prove that Bert's box will not be empty: graphical presentation. The number of coins in his box is a straight line (at 45 degrees). How is that it suddenly drops to 0? And maybe a corollary: Bert never discards more coins that he receives. How is that when he has, let's say 8 coins, he can have less in a later stage? If we reason with {coins} and {events}, don't you see a 1-1 relation?

I was so stuck in my solution that is exactly the same as yours except that the exclusion came earlier that I reacted too quickly. Next time, I will read more carefully, promised

After N steps, they will have received 2*N coins and withdrawn N coins. At that moment, there will be 2*N-N coins in the box. If the box is empty at midnight, this implies: limit(2*N-N)(for N->inf) = 0 At least a little bit surprising. @ThunderCloud I have some troubles to refute your argument. If you remove an infinity of finite numbers from infinity of finite numbers, it does not imply no finite number remain. (Not sure I am convincing and clear enough.) Counterargument: Al removed all coins 1 - N, coins > N remain. If N -> inf, numbering looses it's sense, but he did not remove all coins. As for Charlie, I am ruminating, too. The first idea: every number will remain with p=1/2. Wrong, 1 will be more likely removed than 99. 2nd idea: 1st step, 2 coins: p(removing 1)=1/2 2nd step, 3 coins: p(removing 1)=p(1 was not removed in the first step) * 1/3 = 1/2 * 1/3 = 1/6, p(1 remaining after 2nd step)=1 - 1/2 - 1/6 = 1/3 3rd step, 4 coins: p(removing 1)=p(1 not yet removed) * 1/4 = 1/3 * 1/4 = 1/12, p(1 remaining after 3rd step)=1 - 1/2 - 1/6 - 1/12 = I will not venture further, but this will not converge to 0. (Compare to 1 - 1/2 - 1/4 - 1/8...)

Just a small correction. No one can have 8 coins nor 2. If the total is 21: 5+7+9 is the only possibility. If the total is 4: 1+3+5 is the only possibility

Cannot you shorten it? - if no one else watches A nor B, remove A and B and start over (this happens if A and B are very near and all other are far away enough) - if someone else watches A and/or B, at least one is not being watched (evident; if proof needed, start with 3 and continue with recursion)

Bad news guys, I win: But a mathematician will not agree.

What kind of planes are examined? And what kind of planes are not in the sample? https://en.wikipedia.org/wiki/Abraham_Wald

Real life: I bought (very cheep) a 256MB extension card for a 80286. There were 8 dip switches and no manual. I do not know how many times each switch can be flipped, in any case, it is better to minimize the manipulation. Look https://en.wikipedia.org/wiki/Gray_code

Two judges meet in Moscow, one is laughing, laughing, laughing... The second asks why. 1st judge: I heard an awfully funny joke. 2nd judge: Tell me! 1st judge: I cannot. I sentenced the man who told it to 5 years.

My favorite one: Take the sum of all the integers. Call it S. Take the sum of even integers. That sum is S/2. Because S is infinite, S/2=S, their difference is zero. So the sum of the odd integers is zero.

I googled some definitions of "middle" and "center". If this is THE solution, than the dictionaries are pretty wrong.

Lets try it together. Hidden Content Again, to obvious to be true. But what is wrong? It can be done in a smaller chute. Sure. Applying quantum physics.

One possibility: However, in theory, it can go forever..

Lets try it together. Again, to obvious to be true. But what is wrong?

A typo again...I get used. Thinking it over, it is much more complicated.

So obvious that I fear I am missing something:

@Jasen I think you got it, but SO confusing. 1) Insert the solution on small pieces of paper into the original grid, numbers down. (You better use a non-transparent paper.) 2) For each row/column/square I ask, collect your papers and show them to me in ascending order: (With the original numbers, 1-9 will be used exactly once.) 3) Put your pieces of paper back.

Please delete. 10-11:

@phil1882: The solution does not require a computer. I do not see well encrypting decrypting by hand. All you need: scissors, paper, pencil.

@CaptainEd Good work.... But why so complicated? No need for a third person. <spoiler> 1) The solver secretly creates a matrix with the complete solution. Known numbers are preceded by a star. 2) The challenger writes a program that takes as input this matrix. The program displays numbers preceded by a star and blanks for numbers not preceded by a star and makes the necessary checks. (If the solver fears the program would display everything, it can be tested on another grid.) 3) The solver wipes the harddisk (optional). </spoiler> Almost there. Just the computerized solution does not have the beauty of the manual solution - as I said, it is an intermediate step. How can it be done without a computer? All you need: scissors, paper, pencil.